Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		double[]arr=new double[n];
		double[][]dp=new double[n+1][n+1];
		dp[0][0]=1;
		for(int i=0;i<n;i++){
		    arr[i]=sc.nextDouble();
		}
		for(int j=1;j<=n;j++){
		    dp[j][0]=dp[j-1][0]*(1-arr[j-1]);
		}
		for(int i=1;i<=n;i++){
		    for(int j=1;j<=n;j++)
		        dp[i][j]=arr[i-1]*dp[i-1][j-1]+(1-arr[i-1])*dp[i-1][j];
		}
		double sum=0.0;
		for(int i=n;i>n/2;i--)
		    sum+=dp[n][i];
		System.out.println(sum);
	}
}